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dmitriy555 [2]
2 years ago
7

Which is a better buy. 5 grapfruit for 1.99 or 8 grape fruit for 2.99

Mathematics
1 answer:
11Alexandr11 [23.1K]2 years ago
8 0
8 grape fruits for 2.99
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Bruce drew a scale drawing of a restaurant. The scale he used was 1 millimeter 10 meters The restaurant's kitchen is 3 millimete
Burka [1]

Answer:

answer=30

Step-by-step explanation:

by taking the length as x

1 millimetre = 10 metres

3 millimetres = x

10×3= 30 metres

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What is the probability of getting heads, tails, and heads, in that order, when flipping a coin? Are these events dependent or i
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The answer is 1/8. Now the problem that you gave us isn't dependent upon anything, so more specifically the answer is 1/8 independent.

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6 0
3 years ago
Can pls someone help me with my homework
Wewaii [24]

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4 0
3 years ago
URGENT please help!!! I need a explanation please!! You own a restaurant and can reopen as long as people seated at a table are
hram777 [196]

Answer:

1) Maximum number of tables that can be kept in the dinning room is 10 tables

2)The maximum number of customers, given 6 people per table is 60 customers

Step-by-step explanation:

The parameters given are;

Distance between people seated at different tables = 6 ft

Dimension of table = 2 m by 1.5 m

Number of people at a table = 6 people

Dimension of dinning;

Width = 12 m

Length = 20 m

Dimension of entrance;

Width = 20/21*30  m

Length = 10 m

With 6 meters between tables and a table with of 1.5, we have for the arrangement in the question;

3 tables = 4.5 m

Distance between = 2 × 6 = 12

4.5 + 12 = 16 m (More room required)

With two tables, we have;

Width of tables = 2×1.5 = 3 m

Distance between = 6 m

Dining room width required = 3 + 6 = 9 m

Therefore, the maximum tables in each row = 2

Given that the dining room area extends to the entrance area, total length = 30 m.

With 4 tables we have;

Length = 4 × 2 + 6 × 3 = 26

While on the entrance side of the dinning room area which is 20 m, we have 3 tables;

Length = 3 × 2 + 6 × 2 = 18

Therefore, both arrangements are acceptable;

Just on entering by the right the length = 20/21×30 m

Therefore, with 3 tables will required number due to proximity wit the door and tables arranged along the right wall of the dinning

1) Maximum number of tables that can be kept in the dinning room = 4 + 3 + 3 = 10 tables

2)The maximum number of customers, given 6 people per table = 6×10 = 60 customers.

4 0
3 years ago
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